Question

Find the length of the rectangle given the area is x2-4x-45 ft2 and the width is x+5

Answer 1

`\textit{area of a rectangle}\\\\ A=Lw~~ \begin{cases} L=length\\ w=width\\[-0.5em] \hrulefill\\ A=x^2-4x-45\\ w=x+5 \end{cases}\implies x^2-4x-45=L(x+5) \\\\\\ \cfrac{x^2-4x-45}{x+5}=L\implies \cfrac{(x-9)~~\begin{matrix} (x+5) \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}{~~\begin{matrix}x+5 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~ }=L\implies x-9=L`